CCCC L2-005. 集合相似度

https://www.patest.cn/contests/gplt/L2-005

题解：直接set的count函数

坑 ：要用容斥原理算两个集合的并，否则超时。（我还以为要打表呢）

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
#include <cmath>
#include <cstring>
#include <string>
#include <map>
#include<stack>
#include<set>
#include<string.h>
#include<list>
#define pb push_back
#define mp make_pair
#define _for(i, a, b) for (int i = (a); i<(b); ++i)
#define _rep(i, a, b) for (int i = (a); i <= (b); ++i)
using namespace std;
const  int N = 50 + 5;
set<int> s[N];
int main() {
	int n;
	cin >> n;
	_for(i, 1, n+1) {
		int k;
		cin >> k;
		_for(j, 0, k) {
			int x;
			cin >> x;
			s[i].insert(x);
		}
	}
	int q;
	/*_for(i,1,n)
		_for(j, i+1, n + 1) {
		for (auto t : s[i]) if (s[j].count(i)) tc[i][j]++;
		tt[i][j] = s[i].size() + s[j].size() - 2*tc[i][j];
	}*///写了一半感觉容斥定理可能能ac
	cin >> q;
	_for(i, 0, q) {//2000*4000 8e6 tle;
		int x, y;
		cin >> x >> y;
		float nc=0, nt=0;
		set<int> t;
		for (auto t1 : s[x]) { if (s[y].count(t1)) nc++; }
			//for (auto t2 : s[y]) {t.insert(t2);}
			 nt = s[x].size()+s[y].size()-nc;
			printf("%.2lf%%\n", nc / nt * 100);
	}
	system("pause");
}